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Each should guess whatever would make an even number of red hats. That makes it a 50% chance they're all wrong, and a 50% chance they're all correct -- thus consolidating all their "rightness" into a single block of probability.

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That assumes a uniform distribution. If there were more red hats than blue hats in general, it would be more advantageous for everyone to always guess whatever makes an even number of the majority color, as they would win more than 50% of the time.

I'm not following. With 100 hats, an even number of one color means there's an even number of the other color, too.

Yeah, sorry, I'm mistaken. If one color is more probable than the other, you can win more by guessing odd rather than even, but you'd have to know the distribution. If you got 51 red hats every time, for example, you'd never win making even hats.

For this to work, however, you'd have to know the exact expected probability of each hat (and if it was even or odd), which is improbable.

Indeed, this is the solution. (One can also note that for for the original puzzle of two prisoners, guessing parity reduces to checking whether their hats are the same or different.)

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